Introduction to Relations

Step 1: On division by (3), the possible remainders are (0,1,2). Step 2: Elements with the same remainder go into the same class. Step 3: Hence (3) equivalence classes are formed.

Step 1: Dividing (2) by (3) gives remainder (2). Step 2: Dividing (5) by (3) also gives remainder (2). Step 3: Elements with the same remainder belong to the same equivalence class.

Step 1: Elements placed in the same block are related to each other. Step 2: (1) and (2) are in the same block ({1,2}). Step 3: Therefore ((1,2)) must be in the relation.

Step 1: In an equivalence relation formed by a partition, only elements in the same block are related. Step 2: (1) and (2) are in different blocks. Step 3: Therefore ((1,2)) will not be in the relation.

Step 1: Equivalence relations correspond to partitions of a set. Step 2: For two elements, the partitions are both separate or both together. Step 3: Hence there are (2) equivalence relations.

Step 1: The number of equivalence relations equals the number of partitions. Step 2: For three elements, there are (5) possible partitions. Step 3: Therefore there are (5) equivalence relations.

Step 1: The empty relation has no pair, so there is no pair that can violate symmetry or transitivity. Step 2: But on a non-empty set, diagonal pairs are needed for reflexivity and they are absent. Step 3: Hence it is not reflexive.

Step 1: The universal relation contains all possible pairs, so it contains all diagonal pairs. Step 2: Every reverse pair and every pair required for transitivity is also present. Step 3: Therefore it is an equivalence relation.

Step 1: In a reflexive relation, ((a,a)) is present for every (a\in A). Step 2: Thus every element appears as both first and second component. Step 3: Therefore both domain and range are (A).

Step 1: In a symmetric relation, ((a,b)) comes with ((b,a)). Step 2: Any element appearing as a first component also appears as a second component. Step 3: Hence domain and range are equal.

Step 1: ((1,2)) and ((2,3)) are in the relation. Step 2: Transitivity requires ((1,3)), but it is absent. Step 3: In transitivity check the third pair whenever the middle element matches.

Step 1: All diagonal pairs are present, so the relation is reflexive. Step 2: From ((1,2)) and ((2,3)), ((1,3)) is present, so transitivity is safe. Step 3: The reverse ((2,1)) of ((1,2)) is absent, so it is not symmetric.

Step 1: The relation is reflexive and the given non-diagonal pairs have their reverses. Step 2: But ((1,2)) and ((2,3)) require ((1,3)) by transitivity. Step 3: This pair is missing, so it is not an equivalence relation.

Step 1: (A\times A) contains all possible pairs. Step 2: Reversing any pair still gives a pair in (A\times A). Step 3: Therefore the inverse of the universal relation is the universal relation itself.

Step 1: A reflexive relation contains every ((a,a)). Step 2: Reversing ((a,a)) again gives ((a,a)). Step 3: Therefore the inverse relation also remains reflexive.

Step 1: In a symmetric relation, the reverse of every pair is already in the relation. Step 2: Taking the inverse gives the same set of pairs back. Step 3: A simple identity for a symmetric relation is (R^{-1}=R).

Step 1: In ((1,3)) and ((3,3)), the middle element (3) matches. Step 2: Transitivity requires ((1,3)), which is already present. Step 3: In transitivity, the required pair can sometimes be an already existing pair.

Step 1: (1) and (2) are related because ((1,2)) and ((2,1)) are present. Step 2: (3) is related only to itself. Step 3: Therefore the classes are ({1,2}) and ({3}).

Step 1: (1) is odd and (4) is even. Step 2: In the same-parity relation, both numbers must have the same parity. Step 3: Therefore ((1,4)) is not in the relation.

Step 1: In the given relation, both elements must have the same parity. Step 2: (2) and (4) are both even, so ((2,4)) belongs to the relation. Step 3: In parity relations pair even with even and odd with odd.

Step 1: The domain is formed from first components. Step 2: The first components are (1,2,1,2). Step 3: Removing repetition gives the domain ({1,2}).

Step 1: Reflexivity is checked for every element of the base set (A). Step 2: (3) is in (A), but ((3,3)) is not in the relation. Step 3: Do not decide reflexivity only from the domain; check the whole base set.

Step 1: Every ((a,a)) is present in both (R) and (S). Step 2: The diagonal pairs common to both remain in the intersection. Step 3: Therefore (R\cap S) is reflexive.

Step 1: If a pair is in (R\cup S), it belongs to (R) or (S). Step 2: In that relation, its reverse is also present, so the reverse is in (R\cup S). Step 3: Hence the union of symmetric relations remains symmetric.

Step 1: Pairs coming from different relations can form a new chain. Step 2: The third pair required for that new chain may not be present in the union. Step 3: Therefore the union of transitive relations is not always transitive.